JExocet Compendium

Advanced methods and approaches for solving Sudoku puzzles

Re: JE and Uniqueness

Postby sultan vinegar » Sun Apr 26, 2015 4:20 am

Great work DPB. Some of those eliminations are absolute zingers! My favourite is the STP, I think that is really clever.

Allow me to make my own contribution, apologies if this is old material, but it wasn't covered in your guide. I have a suspicion that it relates to the ABI loop, which is a concept that I never understood!

What better example than the one in which the exocet was born: Fata Morgana.

000000003001005600090040070000009050700050008050402000080020090003500100600000000

This should be straightforward because it is rank 0 logic and there are no triplets. The only caveat is that there is one virtual SIS called V1, which handles the uniqueness constraint:

V1 = {28r3c4, 8r3c6, 7r7c4, 47r7c6}. At least one candidate in V1 must be true to prevent the forbidden pattern of 136 in cells 3n4, 3n6, 5n4, 5n6, 7n4 and 7n6. A little bonus is that our rank 0 elimination also proves that at most one candidate in V1 must be true, giving us a derived weak inference to put up our sleeve for later if required.

Here are the 15 truths and 15 links:

Image

48 Candidates,
15 Truths = {1V1 136R3 136R7 136C2 136C8 5N46}
15 Links = {136r5 4n2 37n4 37n6 6n8 1b37 3b19 6b19}
8 Eliminations --> r15c3<>6, r59c7<>3, r4c2<>24, r2c1<>3, r8c9<>6.

Note that there are no eliminations that haven't been done before, but nonetheless this is a simpler and neater method which may be built upon. Note also that this isn't strictly an exocet. The truths used combine SIS from fish that comprise the traditional exocet plus SIS from the complementary fish. There are obviously deep links between exocets, SK loops and multifish which may be unified into the single grand theory one day.
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Re: JExocet Compendium

Postby David P Bird » Sun Apr 26, 2015 1:47 pm

Hi Sultan Vinegar,

It's good to know that someone appreciates my efforts!

To tackle ABI-loops first. Champagne and I have differing views about what a pattern is. He maintains it is anything that can be found that produces an elimination while I maintain that, once learnt, it has to be recognisable without needing to track the logic. As an alternative to ABI-Loops I therefore defined a pattern to identify incompatible base pairs that would result in a UR being formed. My guess is that it will catch a high proportion of Champagnes eliminations but I've never checked that (covered in document 4).

I appreciate that players such as you using XSudo can delve quite deeply into Exocets, SK loops and Multi-fish but I was pitching my coverage at players who are lower down the learning curve! For instance I only included multi-fish type logic for patterns containing a recognisable SK loop. So, as you say, there is great scope for taking the analysis further.

My criticism of XSudo is that the truth and link sets are reported in a rabbit-out-of the-hat fashion so give little insight into how they were discovered and generally won't qualify as patterns. I therefore don't use it and so am ignorant about how to go about defining a virtual set! But your diagram shows that r357c46 has been reduced to a 6 cell deadly pattern so I suspect there's been a slip between cup and lip.

Here's my simpler and therefore more long-winded start

........3..1..56...9..4..7......9.5.7.......8.5.4.2....8..2..9...35..1..6........ Fata Morgana
Code: Select all
 *--------------------------*--------------------------*--------------------------*
 | 2458    2467    245678   | 126789  16789   1678     | 24589   1248    <3>      | 1.6
 | 2348    2347    <1>      | 23789   3789    <5>      | <6>     248     249      | .3.
 | 2358    <9>     2568     | 12368   <4>     1368     | 258     <7>     125      |
 *--------------------------*--------------------------*--------------------------*
 | 12348   12346 t 2468     | 13678   13678   <9>      | 2347    <5>     12467    |
 | <7>     12346   2469     | 136 b   5       136 b    | 2349    12346   <8>      |
 | 1389    <5>     689      | <4>     13678   <2>      | 379     136 t   1679     |
 *--------------------------*--------------------------*--------------------------*
 | 145     <8>     457      | 1367    <2>     13467    | 3457    <9>     4567     |
 | 249     247     <3>      | <5>     6789    4678     | <1>     2468    2467     | ..6 
 | <6>     1247    24579    | 13789   13789   13478    | 234578  2348    2457     | 13. 
 *--------------------------*--------------------------*--------------------------*
           CL1                        CLb                        CL2


(136)JE2:r5c46,r4c2,r6c8
=> r4c2 <> 24 (non-base digits in target)
(13) & (16) are incompatible base pairs, so (1) can't be a base digit
=> r5c46,r4c2,r6c8 <> 1
(36)NP:r5c46 => r4c45,r5c28,r6c6 <> 36, r5c3 <> 6, r5c7 <> 3
(36)r4c2,r6c8 [JE]=> r4c7 <> 3, r4c7 <> 6, r6c1 <> 3, r6c3 <> 6 (cells see both targets)
(3)r2c5 = (3)r9c5 - (3)r9c8 = (3)r5c8 -[JE]- (3)r4c2 = (3)r2c2 - Loop => r2c14,r9c467 <> 3
(6)r1c5 = (6)r8c5 - (6)r8c8 = (6)r5c8 -[JE]- (6)r4c2 = (6)r1c2 - Loop => r2c346,r8c69 <> 6
(3)r2c5 = (3)r9c5 - (3)r9c8 = (3-6)r6c8 = (6-9)r6c9 = (9)r2c9 => r2c5 <> 9

At this point it gets tougher than I have time for right now. However note that a (36)UR involving the base cells is impossible as one of box 2 & 8 must contain (3) and the other (6) in the base cross-line which is what the incompatibility test effectively checks.

David
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Re: JExocet Compendium

Postby sultan vinegar » Mon Apr 27, 2015 11:50 am

Hi DPB,

I would argue that the pattern I provided is not rabbit-out-of-the-hat at all. I think it will always be there for a (swordfish based, maybe not some of the extensions like the singles free ride etc?) JE that has a pool of just 3 base candidates. Why? Because the green truths are from the fish that comprise the exocet, the purple truths are from the complementary fish (these must "line up" with the base cells otherwise the base cells wouldn't be "in the way" of the fish and we wouldn't have an exocet) and the cell truths are obviously from the base of the exocet.

The trick is the virtual set. There is no need to panic here. It is just a collection of elements at least one of which must be true, just like a native set, except a native set obviously has exactly one true. In r357c46 it is impossible for example to have the combination 1r3c4, 3r3c6, 6r5c4, 1r5c6, 3r7c4 and 6r7c6. Any combination in those six cells solely from the pool of JE base candidates {1,3,6 in this case} is impossible. Why? Well imagine that such a combination was the solution, and every other cell of the puzzle was filled in. There would be two candidates from {1,3,6} available for r3c46, because one of r3c1379 would have been filled in with a 1,3 or 6, and likewise for r5c357 and r7c357. For definiteness let's suppose that we are left with {1,3} in r3c46, {3,6} in r5c46 and {1,6} in r7c46. How do you solve the puzzle from there? There are two possible solutions: {1r3c4, 3r3c6, 3r5c4, 6r5c6, 6r7c4, 1r7c6} and {3r3c4, 1r3c6, 6r5c4, 3r5c6, 1r7c4, 6r7c6}. This is a contradiction under the assumption that the solution is unique, so it must be that such a combination can't be the solution, therefore one of the "uniqueness spoilers", that is, elements of the virtual set must be true.

Now we're home. We have a structure (pattern) with at least 15 true candidates (from the 15 truths), and at most 15 true candidates (from the 15 links) and no triplets, i.e. exactly 15 true candidates (the classic fish principle). Everything in the pattern is rank 0, and we get the bonus of now knowing that exactly one element of the virtual set is true, i.e. the virtual set is both strong and weak. For the manual solver this pattern should be easy to apply because it only needs to be constructed after a JE with a pool of 3 base digits is found. You don't have to go looking for it. As I said, I think that this pattern will always exist for a classic JE with a pool of just 3 base candidates, as everything needed has to be there!

Here it is applied to PlatinumBlonde (one extra truth and link because it is a twin, sorry JE3 in the new nomenclature):

Image

54 Candidates,
16 Truths = {1V1 679R4 4679R7 679C3 679C4 12N7}
16 Links = {679c7 12n3 12n4 47n8 7n9 6b57 7b48 9b48}
13 Eliminations --> r4c8<>234, r6c57<>6, r6c17<>7, r9c57<>9, r7c8<>23, r7c9<>1, r8c6<>7,

Finally, I must confess that I'm not fully across your incompatible base pair heuristics. To me the technique seems to be about what candidates are remaining in a box that might form URs, but I'm not getting the link between that and which band certain candidates lie in. I'm sure something will click on my next pass through your description, which I'm off to right now. Once I understand that then I'll be able to follow your start for FM.

P.S. I like to use Xsudo to find/prove new patterns which I can then use to solve puzzles. Remember that the exocet pattern was found as a rabbit-out-of-the-hat elimination in Xsudo, and it then took a while to understand what was really going on.

SV
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Re: JExocet Compendium

Postby champagne » Mon Apr 27, 2015 12:43 pm

sultan vinegar wrote:P.S. I like to use Xsudo to find/prove new patterns which I can then use to solve puzzles. Remember that the exocet pattern was found as a rabbit-out-of-the-hat elimination in Xsudo, and it then took a while to understand what was really going on.

SV


the exocet pattern came relatively quick after Allan Barker produced the XSUDO eliminations and Allan Barker himself gave the main clues. But you are right, the full power of the pattern came later when "abi" produced the "abi loop" on platinum blonde. This has also been the understanding of that special exocet with a locked candidate
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Re: JExocet Compendium

Postby David P Bird » Mon Apr 27, 2015 3:04 pm

Thanks for your response Sultan. I don't want to engage in an argument about solving methods as it's up to individuals to decide what they feel comfortable with. However one of my concerns is providing solutions that are explainable to others so they too are able to apply the same techniques.
I wrote:My criticism of XSudo is that the truth and link sets are reported in a rabbit-out-of-the-hat fashion so give little insight into how they were discovered and generally won't qualify as patterns.

I want to know the logic that was applied in determining the sets that have been discovered. With the description you've just provided (which I've yet to digest) you've gone a long way to achieving that. In fact XSudo can be a useful validating tool to show that an approach is generic and not just a lucky hit found for a particular circumstance, so I'm not rubbishing it as you seem to believe.

My solving approach is to try to solve puzzles using linear chains supported by recognisable patterns which are the only places where branching can occur. To use an old analogy, to use net methods from the start is like using an elephant gun to exterminate mice. That said following my start of Fata Morgana yesterday, today I find the only way I can find any further eliminations is by resorting to branching.

When you work through my incompatible pairs pattern perhaps you'll find a way to convert it into a truth and link sets including virtual sets to achieve the same eliminations. My hope is then you will find it quicker to look for the pattern elements I describe than to use XSudo or the ABI-loop procedure to reach the same eliminations for further puzzles.

However, running your puzzles through a computer solver, I find that in Fata Morgana r7c6 contains (4) and in Platinum Blonde r1c4 contains (4) both of which you've eliminated [wrong! see following post] so there must be a hole somewhere in your approach. In Platinum Blonde the pattern is (679)JE+ r12c7,r4c8,(4)r7c89 and (6) is found to be incompatible with (7) and (9). It's an interesting puzzle as it contains another sub-pattern I describe in the JE+ document, but even after using that I have to resort to nets to solve it.

David
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Re: JExocet Compendium

Postby David P Bird » Mon Apr 27, 2015 6:59 pm

SV, on reading through your description I see I've been misinterpreting the red candidates in your virtual set as eliminations when they aren't. All your other eliminations are valid, so your model appears to work.

Your eliminations in the JExocet band are the standard ones that can be made as soon as the pattern is identified. But your model also makes base digit eliminations in the parallel bands that are non-standard and would normally require the true base digits to be identified first, and these are interesting.

For the two puzzles you've chosen it's possible to identify incompatible digits and these extra eliminations can be made along with several others. However the compatibility checks aren't always conclusive so these eliminations aren't guaranteed. So the question is under what circumstances are these eliminations of yours available?

David
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Re: JExocet Compendium

Postby daj95376 » Mon Apr 27, 2015 10:34 pm

David P Bird wrote:(136)JE2:r5c46,r4c2,r6c8
=> r4c2 <> 24 (non-base digits in target)
(13) & (16) are incompatible base pairs, so (1) can't be a base digit
=> r5c46,r4c2,r6c8 <> 1
(36)NP:r5c46 => r4c45,r5c28,r6c6 <> 36, r5c3 <> 6, r5c7 <> 3
(36)r4c2,r6c8 [JE]=> r4c7 <> 3, r4c7 <> 6, r6c1 <> 3, r6c3 <> 6 (cells see both targets)
(3)r2c5 = (3)r9c5 - (3)r9c8 = (3)r5c8 -[JE]- (3)r4c2 = (3)r2c2 - Loop => r2c14,r9c467 <> 3
(6)r1c5 = (6)r8c5 - (6)r8c8 = (6)r5c8 -[JE]- (6)r4c2 = (6)r1c2 - Loop => r2c346,r8c69 <> 6
(3)r2c5 = (3)r9c5 - (3)r9c8 = (3-6)r6c8 = (6-9)r6c9 = (9)r2c9 => r2c5 <> 9

Hmmm!!! I missed (never understood) how you determine that a base pair are incompatible. For Fata Morgana, there are two permutations for r5c46=16, and there are two permutations for placing (16) in the target cells. This results in the following four scenarios. Would you please provide, in vivid detail, how you determine that (16) is incompatible in these scenarios.

Code: Select all
Fata Morgana   -plus-   r5c5=5

 r5c46=16; r4c2=1; r6c8=6       r5c46=16; r4c2=6; r6c8=1
 +-----------------------+      +-----------------------+
 | . . . | . . . | . . 3 |      | . . . | . . . | . . 3 |
 | . . 1 | . . 5 | 6 . . |      | . . 1 | . . 5 | 6 . . |
 | . 9 . | . 4 . | . 7 . |      | . 9 . | . 4 . | . 7 . |
 |-------+-------+-------|      |-------+-------+-------|
 | . 1 . | . . 9 | . 5 . |      | . 6 . | . . 9 | . 5 . |
 | 7 . . | 1 5 6 | . . 8 |      | 7 . . | 1 5 6 | . . 8 |
 | . 5 . | 4 . 2 | . 6 . |      | . 5 . | 4 . 2 | . 1 . |
 |-------+-------+-------|      |-------+-------+-------|
 | . 8 . | . 2 . | . 9 . |      | . 8 . | . 2 . | . 9 . |
 | . . 3 | 5 . . | 1 . . |      | . . 3 | 5 . . | 1 . . |
 | 6 . . | . . . | . . . |      | 6 . . | . . . | . . . |
 +-----------------------+      +-----------------------+


 r5c46=61; r4c2=1; r6c8=6       r5c46=61; r4c2=6; r6c8=1
 +-----------------------+      +-----------------------+
 | . . . | . . . | . . 3 |      | . . . | . . . | . . 3 |
 | . . 1 | . . 5 | 6 . . |      | . . 1 | . . 5 | 6 . . |
 | . 9 . | . 4 . | . 7 . |      | . 9 . | . 4 . | . 7 . |
 |-------+-------+-------|      |-------+-------+-------|
 | . 1 . | . . 9 | . 5 . |      | . 6 . | . . 9 | . 5 . |
 | 7 . . | 6 5 1 | . . 8 |      | 7 . . | 6 5 1 | . . 8 |
 | . 5 . | 4 . 2 | . 6 . |      | . 5 . | 4 . 2 | . 1 . |
 |-------+-------+-------|      |-------+-------+-------|
 | . 8 . | . 2 . | . 9 . |      | . 8 . | . 2 . | . 9 . |
 | . . 3 | 5 . . | 1 . . |      | . . 3 | 5 . . | 1 . . |
 | 6 . . | . . . | . . . |      | 6 . . | . . . | . . . |
 +-----------------------+      +-----------------------+

_
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Re: JExocet Compendium

Postby Leren » Tue Apr 28, 2015 1:43 am

daj95376 wrote ; Would you please provide, in vivid detail, how you determine that (16) is incompatible in these scenarios.

At the risk of being labelled a butinsky I can try to explain how I do this. Following a procedure described by Champagne (some years ago) I carry out so called UR tests.

The idea is to assume a solution to the Exocet, solve the grid and see if a contradiction turns up. The most common contradiction (if you are lucky) is that a UR appears in the solution.

For example, if you start off with the first of your grids:

Code: Select all
r5c46=16; r4c2=1; r6c8=6
 +-----------------------+
 | . . . | . . . | . . 3 |
 | . . 1 | . . 5 | 6 . . |
 | . 9 . | . 4 . | . 7 . |
 |-------+-------+-------|
 | . 1 . | . . 9 | . 5 . |
 | 7 . . | 1 5 6 | . . 8 |
 | . 5 . | 4 . 2 | . 6 . |
 |-------+-------+-------|
 | . 8 . | . 2 . | . 9 . |
 | . . 3 | 5 . . | 1 . . |
 | 6 . . | . . . | . . . |
 +-----------------------+

Then using only singles the following partial solution arises :

Code: Select all
 +-----------------------+
 | . 6 . | . . . | . 1 3 |
 | . . 1 | . . 5 | 6 . . |
 | . 9 . | 6 4 1 | . 7 . |
 |-------+-------+-------|
 | . 1 . | . . 9 | . 5 . |
 | 7 . . | 1 5 6 | . . 8 |
 | . 5 . | 4 . 2 | . 6 . |
 |-------+-------+-------|
 | 1 8 . | . 2 . | . 9 . |
 | . . 3 | 5 . . | 1 . . |
 | 6 . . | . 1 . | . . . |
 +-----------------------+

Note the UR (16) r35c46 which proves that scenario false. Doing this for the other three grids (and presumably finding a contradiction of some sort in all cases) proves that (16) is an incompatible pair.

For a Jexocet, in addition to placing the Base and Target cells, you can carry out Swordfish eliminations on your grid to assist you in coming to a quick contradiction.

Well, that's how I do it. I wrote the code some years ago so I'm pretty rusty on the details, but it's worked pretty well, so I'm reasonable sure it's bug free.

Using these so called UR tests I also found that 1 could not occupy Base and Target cells.

Here's a link to Champagne's analysis of Fata Morgana where he shows that 1 is not in the Base and Target cells http://gpenet.pagesperso-orange.fr/UX/Sample8FM/FM_fichiers/FMV2.htm

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Re: JExocet Compendium

Postby champagne » Tue Apr 28, 2015 5:17 am

Leren wrote:
Using these so called UR tests I also found that 1 could not occupy Base and Target cells.

Here's a link to Champagne's analysis of Fata Morgana where he shows that 1 is not in the Base and Target cells http://gpenet.pagesperso-orange.fr/UX/Sample8FM/FM_fichiers/FMV2.htm

Leren


Hi leren,

as you say, this is old stuff, that can also be found within that forum.

To have the full history, it is after the "abi loop" has been discovered that Allan Barker introduced the UR threat in the XSUDO model.

So in fact, the diagram produced by sultan vinegar is in line with that.

The double elimination in the base is very very common with exocets of 3 digits.

Side remark, when the text given by the link was written, the evidence of an exocet with a locked digit had not yet flashed. Platinum Blonde is still described as a "non exocet" pattern;
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Re: JExocet Compendium

Postby David P Bird » Tue Apr 28, 2015 8:01 am

DAJ, the order of the digits in the base cells doesn't affect the compatibility test so your four diagrams reduce to two.

The (16)URs to be avoided are [A]=(16)r35c46, [B]=(16)r37c46 in boxes 2 and 8 respectively.

With (1)r4c2 & (6)r6c8:
the true 'S' cells in the target cross lines are (6)r1c2 and (1)r1c8 which eliminate (16)r1c456 leaving (16)r3c46 as a locked pair and making [A]
the true 'S' cells in column 5 are both in box 8 at (1)r9c5 & (6)r8c5 which disrupts [B] twice over.

With (6)r4c2 & (1)r6c8:
the true 'S' cells in the target cross lines are (1)r9c2 and (6)r8c8 which eliminate (1)r9c456 and (6)r8c56 leaving (16)r7c46 as a locked pair and making [B]
the true 'S' cells in column 5 should both be in box 1 but they share the same cell so this is an impossible combination but otherwise [A] would be disrupted twice over.

Therefore (1) & (6) are incompatible.

When there are dual UR threats the rule is the two true 'S' cells in the target cross lines must lie in different parallel bands. The true 'S' cells in the base cross line will then be forced into different boxes so that each one will disrupt a different UR threat.

Translating this into required pattern elements, for two digits to be compatible they must be able to occupy target cross line 'S' cells in a pair of diagonal boxes. In this case the diagonal box pairs are b1&9 and b3&7.

For double JEs the test should be conducted for each single JE separately as their diagonal boxes will be different.

When there is only one option the target cells each digit would have to occupy will be forced.
There are further notes about identifying when the two threats exist in the write-up.
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Re: JExocet Compendium

Postby sultan vinegar » Tue Apr 28, 2015 12:39 pm

DPB, I get incompatible pairs now! Thanks for yet another great technique.

The symmetry of the fish that comprise the exocet is pertinent here. In FM for example, the fish for number 1 is the opposite polarity (for want of a better word) to the fish for numbers 3 and 6 in the sense that outside of the JE band, the fish for 1 has candidates in the opposite band to the fish for 3 and 6. Thus when for example 1 and 3 are assumed for the targets, the opposite polarity of the two fish conspires to eliminate both 1 and 3 from either box 2 or box 8 (depending on which target takes 1 and which target takes 3) except for the potential UR cells, leaving a forbidden pattern together with the base cells. Considering the polarity of the fish might be a good tactic to use to hone in on what the incompatible pairs might be, rather than having to verify every combination every time? I.e. you should be able to tell straight away from the polarity of the fish in FM that the true base cells must be 3 and 6, because mixing the polarities isn't going to work. Then there is just this one case to verify by calculating the true 'S' cells.

One question for thought. This tactic works because for example candidate 1 is eliminated from r2c456 due to the given in box 1. So if the 'S' cell in r1c8 is true then 1 is eliminated from r1c456, leaving candidate 1 in the potential UR cells r3c46. But for a JE, is it always the case that there will be a void mini-line for each of the base candidates respectively?
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Re: JExocet Compendium

Postby David P Bird » Tue Apr 28, 2015 3:36 pm

SV, I'm very pleased you now understand the compatibility pattern as now some of my earlier comments should now make more sense to you.

you wrote:One question for thought. This tactic works because for example candidate 1 is eliminated from r2c456 due to the given in box 1. So if the 'S' cell in r1c8 is true then 1 is eliminated from r1c456, leaving candidate 1 in the potential UR cells r3c46. But for a JE, is it always the case that there will be a void mini-line for each of the base candidates respectively?

It must eventually work out that way when the third instance of the digit is allocated in the last box in the parallel band. The focus is therefore on preventing the pair of danger cells becoming a locked pair which requires one of the digits to occupy an 'S' cell in each box containing the danger pairs.

Here is your XSudo model for Platinum Blonde translated into an Almost Multi-Sector Locked Set - Champagne could do the same with an almost multi-fish if he wanted.
Code: Select all
    *--------------------------*--------------------------*--------------------------*
    | 35678   34589   679-34   | 679-4   5689    4578     | 679     <1>     <2>      |
    | 15678   14589   679-4    | 679-124 125689  124578   | 679     56789   <3>      |
    | 15678   1589    <2>      | <3>     15689   1578     | <4>     56789   6789     |
    *--------------------------*--------------------------*--------------------------*
234 | 237     2349    <1>      | <8>     236     234      | 23679   679-234 <5>      | b4 79
    | 235     <6>     349      | 124     <7>     12345    | <8>     2349    149      | b5 6
    | 2358-7  23458   347      | 1246    1235-6  <9>      | 123-67  23467   1467     |
    *--------------------------*--------------------------*--------------------------*
123 | 1236    123     <8>      | <5>     1239    1237     | 123679  4679-23 4679-1   | b7 6
    | <9>     123     36       | 127     <4>     1238-7   | <5>     23678   1678     | b8 79
    | <4>     <7>     5        | 129     1238-9  <6>      | 123-9   2389    189      |
    *--------------------------*--------------------------*--------------------------*
                      34         124                        679

Almost MS-NakedSet:(234)r4,(123)r7,(34)c3,(124)c4,(679)c7,(79)b4,(6)b5,(6)b7,(79)b8                  
20 Digit Instances in 19 Covered Cells
20 PEs in 13 Cells =>  234r4c8, 7r6c1, 6r6c5, 67r6c7, 23r7c8, 1r7c9, 7r8c6, 9r9c5, 9r9c7
                              + [ 34r1c3, 4r1c4, 4r2c3, 124r2c4 ]
1 invalid potential elimination confined to r12c34, leaving 13 sure eliminations in 9 cells

The cover sets used for the cells in boxes 4578 are identical to the ones used for SK Loops and it is the eliminations in these boxes that interest me – note that cells r47c34 are occupied by givens.

Other things that stand out are that there are a) only 3 base candidates and b) only 2 pairs of danger cells (that could complete URs with the base digits) which simplify the cover sets needed. This is far from the norm and it remains to be seen if the model can be expanded to cover the other possibilities.

DPB
David P Bird
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Re: JExocet Compendium

Postby daj95376 » Tue Apr 28, 2015 6:32 pm

David P Bird wrote:The (16)URs to be avoided are [A]=(16)r35c46, [_B]=(16)r37c46 in boxes 2 and 8 respectively.

With (1)r4c2 & (6)r6c8:
the true 'S' cells in the target cross lines are (6)r1c2 and (1)r1c8 which eliminate (16)r1c456 leaving (16)r3c46 as a locked pair and making [A]
the true 'S' cells in column 5 are both in box 8 at (1)r9c5 & (6)r8c5 which disrupts [_B] twice over.

With (6)r4c2 & (1)r6c8:
the true 'S' cells in the target cross lines are (1)r9c2 and (6)r8c8 which eliminate (1)r9c456 and (6)r8c56 leaving (16)r7c46 as a locked pair and making [_B]
the true 'S' cells in column 5 should both be in box 2 but they share the same cell so this is an impossible combination but otherwise [A] would be disrupted twice over.

Therefore (1) & (6) are incompatible.

Thanks David for the detailed explanation. I was able to detect [A] after placing 6s and 1s when (1)r4c2 & (6)r6c8. Basically, there was only one acceptable template for each value ... and the DP was forced. When (6)r4c2 & (1)r6c8 were used, I detected that it was an impossible combination after placing 6s and being unable to fully place the 1s. I stopped at this point without realizing that placing the 1s outside of r1c5 would have resulted in [_B]. I had a contradiction without creating the DP. This left me wondering if you had used two different criteria for determining incompatibility for the various scenarios. Thus my question.


Thanks Again & Regards, Danny A. Jones

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daj95376
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Re: JExocet Compendium

Postby daj95376 » Tue Apr 28, 2015 9:44 pm

David P Bird wrote:(136)JE2:r5c46,r4c2,r6c8
=> r4c2 <> 24 (non-base digits in target)
(13) & (16) are incompatible base pairs, so (1) can't be a base digit
=> r5c46,r4c2,r6c8 <> 1
(36)NP:r5c46 => r4c45,r5c28,r6c6 <> 36, r5c3 <> 6, r5c7 <> 3
(36)r4c2,r6c8 [JE]=> r4c7 <> 3, r4c7 <> 6, r6c1 <> 3, r6c3 <> 6 (cells see both targets)
(3)r2c5 = (3)r9c5 - (3)r9c8 = (3)r5c8 -[JE]- (3)r4c2 = (3)r2c2 - Loop => r2c14,r9c467 <> 3
(6)r1c5 = (6)r8c5 - (6)r8c8 = (6)r5c8 -[JE]- (6)r4c2 = (6)r1c2 - Loop => r2c346,r8c69 <> 6
(3)r2c5 = (3)r9c5 - (3)r9c8 = (3-6)r6c8 = (6-9)r6c9 = (9)r2c9 => r2c5 <> 9

Now that I better understand your "incompatible base pairs", there's a (36) scenario to consider. Without covering a multitude of details, suffice it to say that the following is the only acceptable scenario where: (36) are true in the Base and Target cells, we can resolve all 6s/3s/1s, and r9c5=1.

Code: Select all
 +-----------------------------------+
 |  .  6  .  |  .  .  .  |  .  1  3  |
 |  .  .  1  |  .  3  .  |  6  .  .  |
 |  3  .  .  |  6  .  1  |  .  .  .  |
 |-----------+-----------+-----------|
 |  . *3  6  | *1  .  .  |  .  .  .  |
 |  . *1  .  | *3  .  6  |  .  .  .  |
 |  .  .  .  |  .  .  .  |  3  6  1  |
 |-----------+-----------+-----------|
 |  1  .  .  |  .  .  3  |  .  .  6  |
 |  .  .  3  |  .  6  .  |  1  .  .  |
 |  6  .  .  |  .  1  .  |  .  3  .  |
 +-----------------------------------+

Needless to say, there's a UR/DP contradiction in this scenario. Thus, r9c5<>1 follows as an additional elimination for the Fata Morgana puzzle.

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Re: JExocet Compendium

Postby sultan vinegar » Thu Apr 30, 2015 11:13 am

David P Bird wrote:Here is your XSudo model for Platinum Blonde translated into an Almost Multi-Sector Locked Set - Champagne could do the same with an almost multi-fish if he wanted.
Code: Select all
    *--------------------------*--------------------------*--------------------------*
    | 35678   34589   679-34   | 679-4   5689    4578     | 679     <1>     <2>      |
    | 15678   14589   679-4    | 679-124 125689  124578   | 679     56789   <3>      |
    | 15678   1589    <2>      | <3>     15689   1578     | <4>     56789   6789     |
    *--------------------------*--------------------------*--------------------------*
234 | 237     2349    <1>      | <8>     236     234      | 23679   679-234 <5>      | b4 79
    | 235     <6>     349      | 124     <7>     12345    | <8>     2349    149      | b5 6
    | 2358-7  23458   347      | 1246    1235-6  <9>      | 123-67  23467   1467     |
    *--------------------------*--------------------------*--------------------------*
123 | 1236    123     <8>      | <5>     1239    1237     | 123679  4679-23 4679-1   | b7 6
    | <9>     123     36       | 127     <4>     1238-7   | <5>     23678   1678     | b8 79
    | <4>     <7>     5        | 129     1238-9  <6>      | 123-9   2389    189      |
    *--------------------------*--------------------------*--------------------------*
                      34         124                        679

Almost MS-NakedSet:(234)r4,(123)r7,(34)c3,(124)c4,(679)c7,(79)b4,(6)b5,(6)b7,(79)b8                  
20 Digit Instances in 19 Covered Cells
20 PEs in 13 Cells =>  234r4c8, 7r6c1, 6r6c5, 67r6c7, 23r7c8, 1r7c9, 7r8c6, 9r9c5, 9r9c7
                              + [ 34r1c3, 4r1c4, 4r2c3, 124r2c4 ]
1 invalid potential elimination confined to r12c34, leaving 13 sure eliminations in 9 cells

The cover sets used for the cells in boxes 4578 are identical to the ones used for SK Loops and it is the eliminations in these boxes that interest me – note that cells r47c34 are occupied by givens.

Other things that stand out are that there are a) only 3 base candidates and b) only 2 pairs of danger cells (that could complete URs with the base digits) which simplify the cover sets needed. This is far from the norm and it remains to be seen if the model can be expanded to cover the other possibilities.

DPB


I see your 13 eliminations and raise you another four to make it 17, without using any AURs!

Image

19 Truths = {1N7 2N7 4N12567 5N34 6N34 7N12567 8N34 9N4}
19 Links = {23r4 123r7 3c3 12c4 679c7 479b4 46b5 6b7 79b8}
17 Eliminations --> r6c57<>6, r6c17<>7, r9c57<>9, r47c8<>2, r47c8<>3, r2c4<>12, r1c3<>3,
r5c6<>4, r6c2<>4, r7c9<>1, r8c6<>7.

With my (slightly more SK-loop like cover sets in b45) I miss out on the elimination of 4r4c8 though. As you say the eliminations are very interesting. Here it seems that there is the SK-loop like AMSLS in boxes 4,5,7,8 and the ALS in r12c7 which get stitched together and magically the truths balance because of the cells in r47c7. Is this just a freaky coincidence or is there a reason for it, hidden in the puzzle symmetries somewhere?
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